What is x if log(4)x = log(x) 6

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justaguide | College Teacher | (Level 2) Distinguished Educator

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The equation `log_4 x = log_x 6` has to be solved for x.

Use the formula `log_a b = (log_x b)/(log_x a)` which can be used to change the base of any logarithm.

`log_4 x = log_x 6`

Change both the logarithm on both the sides to a common base, for example 10

=> `(log_10 x)/(log_10 4) = (log_10 6)/(log_10 x)`

=> `(log_10 x)^2 = log_10 4*log_10 6`

=> `log_10 x = sqrt(log_10 4*log_10 6)`

=> `x = 10^sqrt(log_10 4*log_10 6)`

=> `x ~~ 4.835`

The solution of the equation is `x = 10^sqrt(log_10 4*log_10 6) ~~ 4.835`

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justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

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In the earlier response a small point has been missed out.

`(log_10 x)^2 = log_10 4*log_10 6` has two solutions.

`log_10 x = +-sqrt(log_10 4*log_10 6)`

The negative value gives another solution of the equation.

`log_10 x = -sqrt(log_10 4*log_10 6)`

=> `x = 10^(-sqrt(log_10 4*log_10 6))`

=> `x ~~ 0.20679`

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aruv | High School Teacher | (Level 2) Valedictorian

Posted on

Let

`log_4 x=log_x 6=y`

`=> log_4 x=y`

`=> 4^y=x`           (i)

Also

`log_x 6=y`

`=> x^y=6 `             (ii)

From (i)

`(4^y)^y=x^y`

`=>4^(y^2)=x^y`              (iii)

From (ii) and (iii), we have

`4^(y^2)=6`

`=> log (4^(y^2))=log6`

`y^2log4=log6`

`y^2=(log6)/(log4)`

`y^2=.77815125/.602059991`

`y^2=1.29248125`

`y=+-1.136873453`

Thus

x=4.835774282 and

x=0.206792116

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tonys538 | Student, Undergraduate | (Level 1) Valedictorian

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The equation `log_4 x = log_x 6` has to be solved.

It is possible to write `log_a b` as `(log_c b)/(log_c a)`

`log_4 x = log_x 6`

`log x/log 4 = log 6/log x`

`(log x)^2 = log 4*log 6`

`log x = +-sqrt(log 4*log 6)`

`x = 10^(sqrt(log 4*log 6))` and `x = 10^-sqrt(log 4*log 6)`

The two solutions of the equation are `10^(sqrt(log 4*log 6))` and `10^-sqrt(log 4*log 6)`

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